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Essential Calculus: Early Transcendentals
Found in: Page 434
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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Short Answer

Determine whether the sequence converges or diverges. If it converges, find the limit.

\({a_n} = \sqrt {\frac{{n + 1}}{{9n + 1}}} \)

The sequence converges and the limit is \(\frac{1}{3}\).

See the step by step solution

Step by Step Solution


A sequence\(\left\{ {{a_n}} \right\}\)has the limit \(L\)and we write \(\mathop {\lim }\limits_{n \to \infty } {a_n} = L\;\;or\;\;{a_n} \to L\)as\(n \to \infty \)if we can make the terms\({a_n}\)as close to\(L\)as we like by taking\(n\)sufficiently large. If \(\mathop {\lim }\limits_{n \to \infty } {a_n}\)exists, we say the sequence converges (or is convergent). Otherwise, we say the sequence diverges (or is divergent).


Consider the sequence\({a_n} = \sqrt {\frac{{n + 1}}{{9n + 1}}} \)

Simplifying we have:

\(\begin{aligned}{a_n} &= \sqrt {\frac{{n + 1}}{{9n + 1}}} \\ &= \sqrt {\frac{{\frac{{n + 1}}{n}}}{{\frac{{9n + 1}}{n}}}} \\ &= \sqrt {\frac{{1 + \frac{1}{n}}}{{9 + \frac{1}{n}}}} \end{aligned}\)

Apply limit

Now evaluate the limit as\(n \to \infty \).

\(\begin{aligned}\mathop {\lim }\limits_{n \to \infty } \sqrt {\frac{{1 + \frac{1}{n}}}{{9 + \frac{1}{n}}}} &= \sqrt {\frac{1}{9}} \\ &= \frac{1}{3}\end{aligned}\)

Which is finite.

Hence the sequence \({a_n} = \sqrt {\frac{{n + 1}}{{9n + 1}}} \)converges and value of limit is \(\frac{1}{3}\).

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